4. A food company is concerned that its 16-ounce can of sliced pears is being overfilled. The quality-control department took a sample of 35 cans and found that the sample mean weight was 16.04 ounces, with a sample standard deviation of 0.08 ounces. Test at a 5% level of significance to see if the mean weight is greater than 16 ounces. What is your conclusion?

Answer:

The cost per gallon is US$ 1.87

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of gallons of gas = 13

Cost of the gallons of gas = US$ 24.31

2. What is the cost per gallon?

Cost per gallon = Cost of the gallons of gas/Number of gallons of gas

Replacing with the real values, we have:

Cost per gallon = 24.31/13

Cost per gallon = 1.87

The cost per gallon is US$ 1.87

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

[tex]\bar X=19.2[/tex] represent the battery life sample mean    

[tex]\sigma=2.5[/tex] represent the population standard deviation    

[tex]n=25[/tex] sample size    

[tex]\mu_o =18[/tex] represent the value that we want to test    

[tex]\alpha[/tex] represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

[tex]p_v[/tex] represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:[tex]\mu = 18[/tex]    

Alternative hypothesis:[tex]\mu \neq 18[/tex]    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:[tex]\mu \leq 18[/tex]    

Alternative hypothesis:[tex]\mu > 18[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

[tex]z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4[/tex]    

4) P-value    

First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=25-1=24[/tex]  

Since is a two tailed test for parts I and II, the p value would be:    

[tex]p_v =2*P(t_{(24)}>2.4)=0.0245[/tex]

And for part III since we have a one right tailed test the p value is:

[tex]p_v =P(t_{(24)}>2.4)=0.0122[/tex]

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the [tex]p_v <\alpha[/tex]. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

Answer:

E. I and III only

Step-by-step explanation:

I. .05

III. one-tailed at .01

Answer:

Step-by-step explanation:

The distance of each lap around pavia park is 1 7/8 miles. Converting

1 7/8 miles to improper fraction, it becomes 15/8 miles.

Ellen rode her bike for 3 1/2 laps before leaving the park. Converting

3 1/2 laps to improper fraction, it becomes 7/2 laps.

The total number of miles that Ellen rode her bike in pavia park would be the product of the distance of each lap and the number of laps that he covered. It becomes

15/8 × 7/2 = 105/16 = 6.5626 miles

Answer: A. 14.02 to 15.98

Step-by-step explanation:

Let [tex]\mu[/tex] denotes the mean waiting time for population.

Given : Sample size : n= 64

Sample mean : [tex]\overline{x}=15[/tex]   (minutes)

Population standard deviation = [tex]\sigma= 4[/tex]

Confidence level : 95%

By z-table , the critical values for 95% confidence = z*=1.96

Confidence interval for population mean : [tex]\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}[/tex]

The 95% confidence interval for the population mean of waiting times will be :

[tex]15\pm (1.96)\dfrac{4}{\sqrt{64}}[/tex]

[tex]15\pm (1.96)\dfrac{4}{8}[/tex]

[tex]15\pm (1.96)(0.5)[/tex]

[tex]15\pm 0.98[/tex]

[tex](15-0.98,\ 15+0.98)=(14.02,\ 15.98)[/tex]

Hence, the 5% confidence interval for the population mean of waiting times is 14.02 to 15.98.

Thus , the correct answer is Option A.

The integral that gives the volume of the solid is

∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.

We have,

To find the volume of the given solid using the general slicing method, we need to integrate the areas of the individual slices perpendicular to the base.

Each cross-section perpendicular to the base and parallel to the diameter is a square.

Let's set up the integral to calculate the volume:

First, let's consider a vertical slice at a distance "y" from the x-axis.

This slice would be a square with a side length "2x," where "x" is the horizontal distance from the y-axis to the rightmost edge of the square.

Since the diameter of the semicircle is on the x-axis and the semicircle is centred on the y-axis, we have a right triangle formed by the radius (11), the distance from the y-axis (x), and the distance from the x-axis (y).

Using the Pythagorean theorem: x² + y² = 11²

Solving for "x": x² = 11² - y²,

so x = √(121 - y²)

Now, the area of the square slice is (side length)² = (2x)² = 4x².

Since the limits of integration are determined by the range of y values, which go from -11 to 11 (the radius of the semicircle), the integral for the volume is:

V = ∫[from -11 to 11] 4x² dy

Substitute the expression for "x" in terms of "y":

V = ∫[from -11 to 11] 4(121 - y²) dy

Simplify:

V = 4 ∫[from -11 to 11] (484 - 4y²) dy

Integrate:

V = 4 [484y - (4/3)y³] | from -11 to 11

V = 4 [484(11) - (4/3)(11)³] - [484(-11) - (4/3)(-11)³]

V = 4 [5344 - (4/3) * 1331] + [5344 + (4/3) * 1331]

V = 14276 + 7119

   = 21395 cubic units

Thus,

The integral that gives the volume of the solid is

∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.

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The volume of the solid is found by integrating the area of the square cross sections, which are determined by the y-coordinate on the semicircle, along the x-axis. The equation for the volume integral is V = ∫ from -11 to 11 of 4(11² - x²) dx.

To find the volume of the solid, we need to integrate the area of the square cross sections along the length of the semicircle. The side length of each square is equal to the y-coordinate at that point (which ranges from -11 to 11). The general form of our semicircle in this position is [tex]y = \sqrt (11^2 - x^2),[/tex] where -11 ≤ x ≤ 11. The equation for the area of each square, then, is  [tex]A(x) = (2y)^2 = 4y^2 = 4(11^2 - x^2).[/tex]

Following the disk method for volumes of revolution, we need to integrate the cross-sectional area along the x-axis, from -11 to 11. So the total volume V is given by V = ∫ from -11 to 11 of A(x) dx = ∫ from -11 to 11 of 4(11² - x²) dx.

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Answer:

There is not enough evidence to support the claim that union membership increased.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 400

p = 11.3% = 0.113

Alpha, α = 0.05

Number of women belonging to union , x = 52

First, we design the null and the alternate hypothesis

[tex]H_{0}: p = 0.113\\H_A: p > 0.113[/tex]

The null hypothesis sates that 11.3% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.

Formula:

[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]

[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

Putting the values, we get,

[tex]z = \displaystyle\frac{0.13-0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}} = 1.073[/tex]

now, we calculate the p-value from the table.

P-value = 0.141636

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to support the claim that union membership increased.

The evidence isn't sufficient enough to support the claim that union membership increased.

This is a statistical measurement used to validate a hypothesis against observed data.

Sample size, n = 400

p = 11.3% = 0.113

Alpha, α = 0.05

Number of women belonging to union   = 52

H₀ : p = 0.113

Hₐ : p  > 0.113

This means 11.3% of U.S. workers belong to union and there was an increase.

p = x / n

z = p - p /(( √p(1-p) /n

 = 53/400 = 0.13

z = p - p /(( √p(1 - p) /n)).

Substitute the values into the equation.

z = 0.13 - 0.113 / ((√0.113(1-0.113)/400)) = 1.073

P-value = 0.141636 from the table which is greater than the significance level, hence we accept the null hypothesis.

The evidence is therefore not sufficient enough to support the claim that union membership increased.

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Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{1117}{10} = 111.7[/tex]

Sum of squares of differences = 642.1

[tex]S.D = \sqrt{\frac{642.1}{49}} = 8.44[/tex]

We are given the following in the question:  

Population mean, μ = 106

Sample mean, [tex]\bar{x}[/tex] = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 106\\H_A: \mu > 106[/tex]

We use one-tailed(right) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135[/tex]

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]

Since,                  

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

Answer:

D. The variable is qualitative because it is an attribute characteristic.

Step-by-step explanation:

In an address, the street name is an attribute of the address.

An attribute is a qualitative variable.

So the correct answer is:

D. The variable is qualitative because it is an attribute characteristic.

The variable is qualitative because it is an attribute characteristic.

Option D is correct

A qualitative variable is a variable whose values are varied by attributes or characteristics. Examples are hair color, course done in school, gender, etc.

A quantitative variable is a variable whose values are varied by actual measurement. Examples are number of odd numbers, number of students in a class, the population of a country, etc.

The description of the given variable is:

Street name of address

This description represents the attribute or characteristic of a location. Therefore, it is a qualitative variable

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Answer:

[tex]P(\bar X<3.83)=0.06117[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(4,1.1)[/tex]  

Where [tex]\mu=4[/tex] and [tex]\sigma=1.1[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case  [tex]\bar X \sim N(4,\frac{1.1}{\sqrt{100}})[/tex]

2) Solution to the problem

We are interested on this probability

[tex]P(\bar X<3.83)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X<3.83)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{3.83-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]

[tex]=P(Z<\frac{3.83-4}{\frac{1.1}{\sqrt{100}}})=P(Z<-1.545)[/tex]

And in order to find this probability we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(Z<-1.545)=0.06117[/tex]

And the excel formula to calculate it would be:

"=NORM.DIST(-1.545,0,1,TRUE)"

The probability is 49.20%

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation.\\\\For\ a\ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]

Given that n = 100, μ = 4 mm, σ = 1.1 mm

For x < 3.83 mm:

[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\z=\frac{3.83-4}{1.1/\sqrt{100} } =-0.0187[/tex]

P(x < 3.83) = P(z < -0.0187) = 0.4920 = 49.20%

From the normal distribution table, the probability that the average thickness of the 100 sheets is less than 3.83 mm is 49.20%

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Answer:

0.1325

Step-by-step explanation:

Weight of the first stock (w1) = .25

Weight of the second stock (w2) = .75

Expected return for the first stock (E(R1)) = .08

Expected return for the second stock (E(R2)) = .15

The expected return of the portfolio is given by the weighted average of the expected return of each stock:

[tex]E(R_p)=w_1*E(R_1)+w_2*E(R_2)\\E(R_p)=0.25*.08 +0.75*.15\\E(R_p)=0.1325[/tex]

The portfolio expected return, E(Rp), is 0.1325

Answer:

Step-by-step explanation:

Let X(t) denote the number of machines breakdown at time t.

The givenn problem follows birth-death process with finite space

S={0, 1, 2, 3} with

[tex] \lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}[/tex]

The birth-death process having balance equations [tex]\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2[/tex]

since, state  rate at which leave = rate at which enter

            0      [tex]\lambda_0P_0=\mu_1P_1[/tex]

             1     [tex](\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0[/tex]

             2   [tex](\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1[/tex]

[tex]P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3[/tex]

Since [tex]\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}[/tex]

a)

Average number not in use equals the mean of the stationary distribution [tex]P_1+2P_2+3P_3=\frac{2136}{751}[/tex]

b)

Proportion of time both repairmen are busy [tex]P_2+P_3=\frac{672}{1522}=\frac{336}{761}[/tex]

The average number of machines not in use is 0.5, and the repairmen are both busy 64% of the time. This has been found under the assumption of exponential distribution for both longevity of machines and repair time. The scenario represents a M/M/2 queue in operations research.

In this scenario, the life of the machines and the repair time are governed by exponential distributions. Exponential distribution is often used to model the amount of time until an event occurs, such as machine failure in this case.

(a) To find the average number of machines not in use, we need to consider the rate of machine breakdown and repair. A machine works an average of 10 hours before failure, which translates to a failure rate of 1/10. A single repairman can fix a machine in an average of 8 hours, meaning a rate of 1/8 per repairman, or 1/4 for two repairmen combined. As there are three machines, the average number of machines in use is the ratio of the arrival rate to the service rate: (1/10) / (1/4) = 2.5 machines. This implies that on average, 0.5 machines are not in use.

(b) Both repairmen will be busy when there are at least two machines that require fixing. The proportion of time in which this is the case is obtained by calculating the probability that the number of machines failed is two or more. This is a problem of queueing theory, in particular an M/M/2 queue. The formula for this probability is P(X >= k) = (1 - rho) * rho^(k) / (1 - rho^(c+1)), where rho = arrival rate / service rate, k = 2, and c = 2 (service channels). Substituting rho = 2.5, we obtain P(X >= 2) = 0.64, meaning that the repairmen are both busy 64% of the time.

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Answer:x = 9

Step-by-step explanation:

The attached photo is that of the given diagram. b represents the angle adjacent 75 degrees.

If line m is parallel to line n, it means that angle b degrees and angle (10x + 15) are corresponding angles. Corresponding angles are equal.

Therefore,

b = 10x + 15

The sum of angles on a straight line is 180 degrees. It means that

b + 75 = 180

b = 180 - 75 = 105

Therefore

10x + 15 = 105

10x = 105 - 15 = 90

x = 90/10 = 9

Answer:

x = 9°

Step-by-step explanation:

105° must be equal to 10x + 15 ° for lines to be parallel.

> 105° = 10x + 15°

> 10x = 90°

> x = 9°

Answer:

b. 1.71

[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]  

Step-by-step explanation:

1) Data given and notation

n=827 represent the random sample taken

X=438 represent the  people that indicate that they would like to see the new show in the lineup

[tex]\hat p=\frac{438}{827}=0.5296[/tex] estimated proportion of people that indicate that they would like to see the new show in the lineup

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:[tex]p \leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(Z>1.71)=0.044[/tex]  

If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.5.  

Answer:

a) [tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex]  b) The 90% confidence interval is given by (17.3;21.1)  

c) No, because 18 is above the lower limit of the confidence interval.

d) Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Step-by-step explanation:

1) Notation and definitions  

n=17 represent the sample size

Part a  

[tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex]  Part b

[tex]s=4.4[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

2) Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=17-1=16[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.05,16)" for [tex]t_{\alpha/2}=-1.75[/tex]  

"=T.INV(1-0.05,16)" for [tex]t_{1-\alpha/2}=1.75[/tex]  

The critical value [tex]tc=\pm 1.75[/tex]  

3) Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.75 \frac{4.4}{\sqrt{17}}=1.868[/tex]  

4) Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]19.2 - 1.75 \frac{4.4}{\sqrt{17}}=17.332[/tex]  

[tex]19.2 + 1.75 \frac{4.4}{\sqrt{17}}=21.068[/tex]  

The 90% confidence interval is given by (17.332;21.068)  and rounded would be:  (17.3;21.1)

Part c

No, because 18 is above the lower limit of the confidence interval.

Part d

Because the parent population is assumed to be normally distributed.

The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.

Green's theorem says the circulation of [tex]\vec F[/tex] along the rectangle's border [tex]C[/tex] is equal to the integral of the curl of [tex]\vec F[/tex] over the rectangle's interior [tex]D[/tex].

Given [tex]\vec F(x,y)=2xy\,\vec\imath[/tex], its curl is the determinant

[tex]\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x[/tex]

So we have

[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}[/tex]

The circulation of the vector field 2xyi around the given rectangle, as computed via Green's Theorem, is 0 due to the curl of F being 0.

To use Green's Theorem to calculate the circulation around a rectangle, first we should realize that Green's Theorem states that the line integral around a simple closed curve C of F.dr is equal to the double integral over the region D enclosed by C of the curl of F. Here, F is the vector field defined as 2xyi. The given rectangle is oriented counterclockwise and the values of x and y are given as 0≤x≤8 and 0≤y≤3 respectively. The line integral denotes the circulation of the field.

The circulation is thus the double integral over the rectangle of ∇ x F. But in this case, since F = 2xyi, we get ∇ x F = 0. Hence, the circulation of F around the given rectangle is 0.

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Answer:

Step-by-step explanation:

Given that Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14.

                                      Group I  X             Group II Y

Sample mean                     74                          74

n                                         25                          64

Std error (14/sqrtn)             2.8                        1.75

a) P(X>80) =[tex]1-0.9839\\= 0.0161[/tex]

b) P(Y>80) = [tex]1-0.9997\\=0.0003[/tex]

c) X-Y is Normal with mean = 0 and std deviation = [tex]\sqrt{2.8^2+1.75^2} \\=3.302[/tex]

P(X-Y>2.2) = [tex]1-0.8411\\=0.1589[/tex]

d) [tex]P(\bar x -\bar Y>2.2) = 0.1589[/tex]

Answer: b. (.54, .71)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

,where [tex]\hat{p}[/tex] = sample proportion

z= Critical z-value

n= sample size.

Let p be the proportion of people who support additional handgun control.

As per given , we have

n= 80

[tex]\hat{p}=\dfrac{50}{80}=0.625[/tex]

Critical z-value for 90% confidence interval is 1.645

Now , a 90% confidence interval for the population proportion of those who support additional handgun control will become:

[tex]0.625\pm (1.645)\sqrt{\dfrac{0.625(1-0.625)}{80}}[/tex]

[tex]=0.625\pm (1.645)\sqrt{0.0029296875}[/tex]

[tex]=0.625\pm 0.089\\\\=(0.625-0.089, 0.625+0.089)\\\\=(0.536,\ 0.714)\approx(0.54,\ 0.71)[/tex]

So the correct answer is : b. (.54, .71)

Final answer:

The degrees of freedom for the t-statistic are 18. The conclusion cannot be determined without the p-value. If the null hypothesis is true, there is not enough evidence to support the alternative hypothesis.

Explanation:

(a) The degrees of freedom for the t-statistic are found by subtracting 1 from the sample size. In this case, the sample size is 19 so the degrees of freedom would be 19 - 1 = 18.

(b) If the p-value is less than the alpha level (typically 0.05), we reject the null hypothesis. In this case, the p-value is not provided, so we cannot determine the conclusion.

(c) If the null hypothesis is true, we would not reject it and conclude that there is not enough evidence to support the alternative hypothesis.

Answer:

Step-by-step explanation:

Since the human body temperatures are normally distributed, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = human body temperatures

u = mean body temperature

s = standard deviation

From the information given,

u = 98.20oF

s = 0.62oF

We want to find the probability that their mean body temperature will be less than 98.50oF. It is expressed as

P(x lesser than 98.50)

For x = 98.50,

z = (98.50 - 98.20)/0.62 = 0.48

Looking at the normal distribution table, the corresponding probability to the z score is 0.6844

P(x lesser than 98.50) = 0.6844

To find the closest point on a line to another point, convert the problem into finding a minimum of a distance function, then use calculus (derivatives) to find the minimum.

To solve this problem, you need to identify a goal and an approach using calculus. The goal is to find a point (x, y) on the given line 5x + y = 7 that is closest to the point (−3, 2). This translates into minimizing the distance between (x, y) and (-3, 2) with respect to (x, y).

The distance formula between two points (x1, y1) and (x2, y2) is: √[(x2-x1)²+(y2-y1)²]. For our purposes x1 = -3, y1 = 2 and x2 = x, y2 = y. However, since y is from our given line (5x + y = 7), we can substitute for y = 7 - 5x. Thus, our new distance will become a function in terms of x: d(x) = √[(x - -3)²+(7 - 5x - 2)²].

To minimize it, we should take the derivative of this function, and set it equal to zero, which will give an extreme point (either minimum or maximum). Using the chain-rule results in the minimizing x value, and substituting it back to the line gives us the y value. This resultant point (x, y) would be our answer to the problem.

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Answer:

Step-by-step explanation:

move constant to the right and change the sign

X=12-8

X=4

In this exercise we have to use the knowledge of variance to calculate the value of n, so we have that:

the sample is n=306

Organizing the information given in the statement we have that:

So given by the equation we have:

[tex]Z' = t(0.075)= 1.44\\n = (Z'*S/E)^2\\n = ( 1.44 * 0.85/0.07)^2\\n = (17.4857)^2\\n = 305.75\\n = 306[/tex]

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The minimum number of males over age 25 they must include in their sample is 512.

Given, Desired confidence level: 85%

Maximum error (E): 0.07 liters

Variance ([tex]\sigma^{2}[/tex]): 1.21 liters

Standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(\sigma\)[/tex] = [tex]\sqrt{1.21}[/tex] = 1.1

Z-value for 85% confidence level (lookup Z-value for 0.425 in the standard normal distribution): [tex]\[ Z \approx 1.44 \][/tex]

n= [tex]\left(\frac{Z \sigma}{E}\right)^2[/tex]

[tex]\[ n = \left(\frac{1.44 \times 1.1}{0.07}\right)^2 \][/tex]

n= (1.584/0.07)² = 511.986

n = 512

We can see here that at the 0.01 significance level, we can actually conclude that the single-earner couples on average spend more time watching television together.

To determine whether we can conclude that single-earner couples spend more time watching television together on average than dual-earner couples, we can perform a hypothesis test.

The null hypothesis (H₀) assumes that there is no difference in the mean time spent watching television between the two groups, while the alternative hypothesis (H₁) suggests that single-earner couples spend more time together watching television.

Let's set up the hypotheses:

Null Hypothesis (H₀): μ₁ ≤ μ₂ (The mean time spent together watching television for single-earner couples is less than or equal to the mean time for dual-earner couples.)

Alternative Hypothesis (H₁): μ₁ > μ₂ (The mean time spent together watching television for single-earner couples is greater than the mean time for dual-earner couples.)

Where:

μ₁ = population mean time spent watching television for single-earner couples

μ₂ = population mean time spent watching television for dual-earner couples

Next, we will use a two-sample t-test to test the hypotheses. Since we are trying to determine if single-earner couples spend more time watching television, this will be a one-tailed t-test.

Given the sample means, sample standard deviations, and sample sizes, we can calculate the t-statistic and compare it to the critical t-value at the 0.01 significance level (α = 0.01) with degrees of freedom d f = n₁ + n₂ - 2, where n₁ and n₂ are the sample sizes of single-earner and dual-earner couples, respectively.

Let's assume the sample sizes are n₁ = n₂ = 30 (the actual sample sizes from the study are not given in the question, but this is just for demonstration purposes).

Now, we can calculate the t-statistic:

t = (x₁ - x₂) / √((s₁²/n₁) + ([tex]s_{2}[/tex]² /n₂))

where:

x₁ = sample mean time for single-earner couples

x₂ = sample mean time for dual-earner couples

s₁ = sample standard deviation for single-earner couples

[tex]s_{2}[/tex] = sample standard deviation for dual-earner couples

n₁ = sample size for single-earner couples

n₂ = sample size for dual-earner couples

Using the provided values:

x₁ = 61 minutes

x₂ = 48.4 minutes

s₁ = 15.5 minutes

= 18.1 minutes

n₁ = n₂ = 30 (sample sizes assumed for demonstration)

Calculating the t-statistic:

t = (61 - 48.4) / √((15.5²/30) + (18.1²/30))

t ≈ 4.083

Next, we need to find the critical t-value from the t-distribution table at α = 0.01 significance level and df = 30 + 30 - 2 = 58 (degrees of freedom).

The critical t-value at α = 0.01 with d  f = 58 is approximately 2.660.

Since the calculated t-statistic (4.083) is greater than the critical t-value (2.660), we reject the null hypothesis (H₀).

Therefore, at the 0.01 significance level, we can conclude that single-earner couples, on average, spend more time watching television together than dual-earner couples based on the data provided in the study.

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To determine whether single-earner couples spend more time watching television together than dual-earner couples, an independent samples t-test must be conducted at the 0.01 significance level. Using the provided means and standard deviations, we would calculate a t-statistic and compare it against critical values to either reject the null or fail to reject it.

The question asks whether single-earner couples spend more time watching television together than dual-earner couples, based on a study with provided mean values and standard deviations for both groups. To determine if there is a statistically significant difference between the two means, we would conduct a hypothesis test, specifically an independent samples t-test, at the 0.01 significance level. The null hypothesis (0) would state that there is no difference in the mean television watching times between the groups, while the alternative hypothesis (A) would claim that there is a difference, specifically that single-earner couples watch more television.

Given that the mean time spent watching television for single-earner couples is 61 minutes with a standard deviation of 15.5 minutes, and for dual-earner couples it is 48.4 minutes with a standard deviation of 18.1 minutes, we would calculate the t-statistic and compare it against the t-distribution critical values for the given degrees of freedom. If the calculated t-statistic exceeds the critical value for a one-tailed test at the 0.01 level, we would reject the null hypothesis and conclude that there is a significant difference, supporting the claim that single-earner couples spend more time watching television together.

Answer:

a) [tex]P(t) = 2e^{0.275t}[/tex]

b) 54.225 square centimeters.

c) 2.52 hours

Step-by-step explanation:

The population growth law is:

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, as a decimal.

In this problem, we have that:

The colony occupied 2 square centimeters initially, so [tex]P_{0} = 2[/tex]

The colony triples every 4 hours. So

[tex]P(4) = 3P_{0} = 6[/tex]

(a) An expression for the size P(t) of the colony at any time t.

We have to find the value of r. We can do this by using the P(4) equation.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]6 = 2e^{4r}[/tex]

[tex]e^{4r} = 3[/tex]

Applying ln to both sides, we get:

[tex]4r = 1.1[/tex]

[tex]r = 0.275[/tex]

So

[tex]P(t) = 2e^{0.275t}[/tex]

(b) The area occupied by the colony after 12 hours.

[tex]P(t) = 2e^{0.275t}[/tex]

[tex]P(12) = 2e^{0.275*12}[/tex]

[tex]P(12) = 54.225[/tex]

(c) The doubling time for the colony?

t when [tex]P(t) = 2P_{0} = 2*2 = 4[/tex].

[tex]P(t) = 2e^{0.275t}[/tex]

[tex]4 = 2e^{0.275t}[/tex]

[tex]e^{0.275t} = 2[/tex]

Applying ln to both sides

[tex]0.275t = 0.6931[/tex]

[tex]t = 2.52[/tex]

Answer:

C. There is not enough information to answer this question

Step-by-step explanation:

Conclusion cannot be made whether to retain or reject the null hypothesis because the information given is not sufficient

Answer:

There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a UH student checks out books on history.

B is the probability that a UH students checks out books on science.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that a UH student checks a book on history but not on science and [tex]A \cap B[/tex] is the probability that a UH student checks books both on history and science.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

What is the probability that a randomly selected UH student checks out a history book or a science book or both?

[tex]P = a + b + (A \cap B)[/tex]

We start finding these values from the intersection.

6% check out books on both history and science. So [tex]A \cap B = 0.06[/tex]

28% of all UH students check out books on science. So [tex]B = 0.28[/tex]

[tex]B = b + (A \cap B)[/tex]

[tex]0.28 = b + 0.06[/tex]

[tex]b = 0.22[/tex]

12% of all UH students check out books on history

[tex]A = a + (A \cap B)[/tex]

[tex]0.12 = a + 0.06[/tex]

[tex]a = 0.06[/tex]

So

[tex]P = a + b + (A \cap B) = 0.06 + 0.22 + 0.06 = 0.34[/tex]

There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.

Answer:

Step-by-step explanation:

Given that a sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability

a) The two qualitative variables are disability and not having disability and secondly the grades of children

b) Contingency table:

Grade                  4                5                      Total

Normal read.       32              34                        66

Not normal read.  23                6                        29  

Total                      55             40                         95

H0: Reading disability is independent of grade.

Ha: There is association between the two

c)  4 5 Total

Nor read 38.21052632 27.78947368 66

Not norm 16.78947368 12.21052632 29

Expected cells are obtained using the formula

row total*col total/grand total

Answer:

Option A)  k = 2

Step-by-step explanation:

Multimonial Experiment

A multimonial experiment is an experiment with n repeated trials and each trial has a discrete number of possible outcomes.

Binomial Experiment

Binomial experiment is an experiment with n repeated trials and each trial has only two possible outcomes.

Thus, if k  represents the number of possible outcomes, then for k = 2, a multimonial experiment will become a binomial experiment.

Option A)  k = 2

Answer:

Step-by-step explanation:

If a number is evenly divisible by another number, it means that the number divides it completely without a remainder.

6) 44 is evenly divisible by 2 and 4. 44 divided by 3 and 6 would have remainders.

7) 38 is evenly divisible by 2. There would be remainders if 38 divides 3, 4 or 6

8) 726 is evenly divisible by 2, 3 and 6. It is not evenly divisible by 4

9) 2112 is evenly divisible by 2, 3 4 and 6

10) 1221 is evenly divisible by 3. It is not evenly divisible by 2 4 and 6

Answer:

A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

[tex]4 = \frac{c_{1}}{1 + c_{1}}[/tex]

[tex]4 + 4c_{1} = c_{1}[/tex]

[tex]c_{1} = -\frac{4}{3}[/tex]

A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].

Through the point (0, 1)?

Same thing as above

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

[tex]1 = \frac{c_{1}}{1 + c_{1}}[/tex]

[tex]1 + c_{1} = c_{1}[/tex]

[tex]0c_{1} = 1[/tex]

No solution.

So there is not a solution curve passing through the point(0,1).